JEE Main 2026 Apr 6 Shift 2, Chemistry Q9: Valence Shell Electron Pair Repulsion (VSEPR) Theory

JEE Main2026Apr 6, Shift 2Chemistry

Given below are two statements:

Statement I: F $_{2}$ O < H $_{2}$ O < Cl $_{2}$ O is the correct trend in terms of bond angle.

Statement II: SiF $_{4}$ , SnF $_{4}$ and PbF $_{4}$ are ionic in nature.

In the light of the above statements, choose the correct answer from the options given below:

A
Both Statement I and Statement II are true
B
Both Statement I and Statement II are false
C
Statement I is true but Statement II is false
D
Statement I is false but Statement II is true

Solution

Statement I. Reported bond angles: F $_{2}$ O $\approx 10 3^{\circ}$ , H $_{2}$ O $\approx 104. 5^{\circ}$ , Cl $_{2}$ O $\approx 11 1^{\circ}$ . The substituents on oxygen exert different effects — high electronegativity of F pulls bond pairs away from O (smaller angle), while bulky Cl forces a wider angle. Trend matches the statement. True.

Statement II. SiF $_{4}$ is essentially a covalent molecular fluoride. SnF $_{4}$ and PbF $_{4}$ have significant ionic character but SiF $_{4}$ does not. The blanket claim that all three are ionic is false.

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q9: Valence Shell Electron Pair Repulsion (VSEPR) Theory

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